rsarashi wrote:How many of the integers that satisfy the inequality (x+2)(x+3)/(x-2) >= 0 are less than 5?
A. 1
B. 2
C. 3
D. 4
E. 5
OAD
Hi rsarashi,
Let's observe the inequality (x+2)(x+3)/(x-2) >= 0 and derive results.
1. Since @ x= 2, the denominator (x-2) would become 0, making the inequality undefines, it is not a solution.
2. We see that @ x = 3 and @ x = 4, the numerator (x+2)(x+3), as well as the denominator (x-2) would remain positive, thus x = 3 and x= 4 are two solutions.
3. Let's focus on x < 2 values. Since @ x = 1 or 0 or -1, the numerator (x+2)(x+3) remain positive, the denominator (x-2) becomes negative, making the inequality negative, which is not we want. The inequality >= 0. Thus, x = 1 or 0 or -1 are not the solutions.
4. We see that @ x= -2 and @ x= -3, inequality turns 0, thus these two are the solutions.
5. We must not focus on x < -3 since @x<-3, the numerator (x+2)(x+3) remain positive (negative*negative = positive), the denominator (x-2) becomes negative, making the inequality negative, which is not we want.
So, there are four numbers of integer solutions that are less than 5: x = -3, -2, 3 and 4.
The correct answer:
D
Hope this helps!
Relevant book:
Manhattan Review GMAT Math Essentials Guide
-Jay
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